1. 1. IIT–JEE Syllabus
Atomic structure; Rutherford Model; Spectrum of hydrogen atom; Bohr model;
de Broglie relations, Uncertainty principle, Quantum model; Electronic configuration
of elements ( upto to atomic number 36); Aufbau principle, Pauli’s exclusion principle
and Hund’s rule, shapes of s,p, and d orbitals.
2. Dalton’s Atomic Theory
All the objects around you, this book, your pen or pencil and things of nature such as rocks,
water and plant constitute the matter of the universe. Matter is any substance which
occupies space and has mass.
Dalton, in 1808, proposed that matter was made up of extremely small, indivisible particles
called atoms. (In Greek atom means which cannot be cut). This concept was accepted for
number of years.
The main postulates of Dalton’s atomic theory are
 Matter is made up of small indivisible particles, called atoms.
 Atoms can neither be created nor destroyed. This means that a chemical reaction is
just a simple rearrangement of atoms and the same number of atoms must be present
before and after the reaction.
 Atom is the smallest particle of an element which takes part in a chemical reaction.
 Atoms of the same element are identical in all respects especially, size, shape and
mass.
 Atoms of different elements have different mass, shape and size.
 Atoms of different elements combine in a fixed ratio of small whole numbers to form
compound atoms, called molecules.
However, the researches done by various eminent scientists and the discovery of
radioactivity have established beyond doubt, that atom is not the smallest indivisible particle
but had a complex structure of its own and was made up of still smaller particles like
electrons, protons, neutrons etc. At present about 35 different subatomic particles are
known but the three particles namely electron, proton and neutron are regarded as the
fundamental particles.
We shall now take up the brief study of these fundamental particles. The existence of
electrons in atoms was first suggested, by J.J. Thomson, as a result of experimental work on
the conduction of electricity through gases at low pressures and high voltage, which
produces cathode rays consisting of negatively charged particles, named as electrons. The
e/m ratio for cathode rays is fixed whose values is 1.76  108
C/g
We know that an atom is electrically neutral, if it contains negatively charged electrons it
must also contain some positively charged particles. This was confirmed by Goldstein in his
discharge tube experiment with perforated cathode. On passing high voltage between the

2. electrodes of a discharge tube it was found that some rays were coming from the side of the
anode which passed through the holes in the cathode. These anode rays (canal rays)
consisted of positively charged particles formed by ionization of gas molecules by the
cathode rays. The charge to mass ratio ( e/m value) of positively charge particles was found
to be maximum when the discharge tube was filled with hydrogen gas as hydrogen is the
lightest element. These positively charged particles are called protons.
e/m varies with the nature of gas taken in the discharge tube. The positive particles are
positive residues of the gas left when the gas is ionized.
The neutral charge particle, neutron was discovered by James Chadwick by bombarding
boron or beryllium with –particles.
Characteristics of the three fundamental particles are:
Electron Proton Neutron
Symbol e or e–
p n
Approximate relative mass 1/1836 1 1
Approximate relative charge –1 +1 No charge
Mass in kg 9.10910–31
1.67310–27
1.67510–27
Mass in amu 5.48510–4
1.007 1.008
Actual charge (coulomb) 1.60210–19
1.60210–19
0
Actual charge (e.s.u.) 4.8  10–10
4.8  10–10
0
The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 6C12
, i.e.,
1.66010–27
kg.
The neutron and proton have approximately equal masses of 1 amu and the electron is
about 1836 times lighter, its mass can sometimes be neglected as an approximation.
The electron and proton have equal, but opposite, electric charges while the neutron is not
charged.
3. Atomic Models
We know the fundamental particles of the atom. Now let us see, how these particles are
arranged in an atom to suggest a model of the atom.
3.1 Thomson’s Model
J.J. Thomson, in 1904, proposed that there was an equal and opposite positive charge
enveloping the electrons in a matrix. This model is called the plum – pudding model after a
type of Victorian dissert in which bits of plums were surrounded by matrix of pudding.

3. electron
Positive sphere
This model could not satisfactorily explain the results of scattering experiment carried out by
Rutherford who worked with Thomson.
3.2 Rutherford’s Model
– particles emitted by radioactive substance were shown to be dipositive Helium ions
(He++
) having a mass of 4 units and 2 units of positive charge.
Rutherford allowed a narrow beam of –particles to fall on a very thin gold foil of thickness
of the order of 0.0004 cm and determined the subsequent path of these particles with the
help of a zinc sulphide fluorescent screen. The zinc sulphide screen gives off a visible flash
of light when struck by an  particle, as ZnS has the remarkable property of converting
kinetic energy of  particle into visible light. [For this experiment, Rutherford specifically
used  particles because they are relatively heavy resulting in high momentum].
Observation
i) Majority of the –particles pass straight through the gold strip with little or no deflection.
ii) Some –particles are deflected from their path and diverge.
iii) Very few –particles are deflected backwards through angles greater than 90.
iv) Some were even scattered in the opposite direction at an angle of 180 [ Rutherford was
very much surprised by it and remarked that “It was as incredible as if you fired a 15–
inch shell at a piece of tissue paper and it came back and hit you”]. There is far less
difference between air and bullet than there is between gold atoms and -particles
assuming of course that density of a gold atom is evenly distributed.
Conclusions
1. The fact that most of the  - particles passed straight through the metal foil indicates the
most part of the atom is empty.

4. 2. The fact that few  - particles are deflected at large angles indicates the presence of a
heavy positively charge body i.e., for such large deflections to occur  - particles must
have come closer to or collided with a massive positively charged body.
3. The fact that one in 20,000 have deflected at 180° backwards indicates that volume
occupied by this heavy positively charged body is very small in comparison to total
volume of the atom.
Atomic model
On the basis of the above observation, and having realized that the rebounding -particles
had met something even more massive than themselves inside the gold atom, Rutherford
proposed an atomic model as follows.
i) All the protons (+ve charge) and the neutrons (neutral charge) i.e nearly the total mass
of an atom is present in a very small region at the centre of the atom. The atom’s central
core is called nucleus.
ii) The size of the nucleus is very small in comparison to the size of the atom. Diameter of
the nucleus is about 10–13
cm while the atom has a diameter of the order of 10–8
cm. So,
the size of atom is 105
times more than that of nucleus.
ii) Most of the space outside the nucleus is empty.
iv) The electrons, equal in number to the net nuclear positive charge, revolve around the
nucleus with fast speed in various circular orbits.
v) The centrifugal force arising due to the fast speed of an electron balances the
coulombic force of attraction of the nucleus and the electron remains stable in its path.
Thus according to him atom consists of two parts (a) nucleus and (b) extra nuclear part.
Defects of Rutherford’s atomic model
1. Position of electrons: The exact positions of the electrons from the nucleus are not
mentioned
2. Stability of the atom: Neils Bohr pointed out that Rutherford’s
atom should be highly unstable. According to the law of
electro–dynamics, the electron should therefore, continuously
emit radiation and lose energy. As a result of this a moving
electron will come closer and closer to the nucleus and after
passing through a spiral path, it should ultimately fall into the
nucleus.
+
It was calculated that the electron should fall into the nucleus in less than 10–8
sec. But it is
known that electrons keep moving outside the nucleus.
To solve this problem Neils Bohr proposed an improved form of Rutherford’s atomic model.
Before going into the details of Neils Bohr model we would like to introduce you some
important atomic terms.

5. Atomic terms
a) Atomic Number (Z): The atomic number of an element is the number of protons
contained in the nucleus of the atom of that element.
b) Nucleons: Protons and neutrons are present in a nucleus. So, these fundamental
particles are collectively known as nucleons.
c) Mass Number (A): The total number of protons and neutrons i.e, the number of
nucleons present in the nucleus is called the mass number of the element.
d) Nuclide: Various species of atoms in general. A nuclide has specific value of atomic
number and mass number.
IUPAC notation of an atom (nuclide)
Let X be the symbol of the element. Its atomic number be Z and mass number be A.
Then the element can be represented as A
Z X
e) Isotopes: Atoms of the element with same atomic number but different mass number
e.g. 1H1
, 1H2
, 1H3
. There are three isotopes of hydrogen.
f) Isobars: Atoms having the same mass number but different atomic numbers, e.g. 15P32
and 16S32
are called isobars.
g) Isotones: Atoms having the same number of neutrons but different number of protons or
mass number, e.g. 6C14
, 8O16
, 7N15
are called isotones.
h) Isoelectronic: Atoms, molecules or ions having same number of electrons are
isoelectronic e.g. N2,CO, CN–
.
i) Nuclear isomer: Nuclear isomers (isomeric nuclei) are the atoms with the same atomic
number and same mass number but with different radioactive properties.
Example of nuclear isomers is
Uranium –X (half life 1.4 min) and
Uranium –Z (half life 6.7 hours)
The reason for nuclear isomerism is the different energy states of the two isomeric
nuclei.
Other examples are
30Zn69
30Zn69
(T1/2 = 13.8 hr) (T1/2 = 57 min)
35Br80
35Br80
(T1/2 = 4.4 hour) (T1/2 = 18 min)
j) Isosters : Molecules having same number of atoms and also same number of electrons
are called isosters.
E.g., (i) N2 and CO
ii) CO2 and N2O
iii) HCl and F2
k) Atomic mass unit: Exactly equal to 1/12 of the mass of 6C12
atom
1 amu = 1.66 10–27
kg = 931.5 MeV

6. 4. Some Important Characteristics of a Wave
A wave is a sort of disturbance which originates from some vibrating source and travels
outward as a continuous sequence of alternating crests and troughs. Every wave has five
important characteristics, namely, wavelength (), frequency (), velocity (c), wave number
 
 and amplitude (a).
a
 Crest
Crest

Trough
Trough
Ordinary light rays, X–rays,–rays, etc. are called electromagnetic radiations because similar
waves can be produced by moving a charged body in a magnetic field or a magnet in an
electric field. These radiations have wave characteristics and do not require any medium for
their propagation.
i) Wave length (): The distance between two neighbouring troughs or crests is known as
wavelength. It is denoted by  and is expressed in cm, m, nanometers (1nm=10–9
m) or
Angstrom (1Å=10–10
m).
ii) Frequency (): The frequency of a wave is the number of times a wave passes through
a given point in a medium in one second. It is denoted by (nu) and is expressed in
cycles per second (cps) or hertz (Hz) 1Hz = 1cps.
The frequency of a wave is inversely proportional to its wave length ()
 

1
or  =

c
iii) Velocity: The distance travelled by the wave in one second is called its velocity. It is
denoted by c and is expressed in cm sec–1
.
c =  or  =

c
iv) Wave number  
 : It is defined as number of wavelengths per cm. It is denoted by
 and is expressed in cm–1
.
 =

1
(or)  =
c

v) Amplitude: It is the height of the crest or depth of the trough of a wave and is denoted
by a. It determines the intensity or brightness of the beam of light.

7. Wavelengths of electromagnetic radiations
Electromagnetic radiations Wave length (Å)
Radio waves 31014
to 3 107
Micro waves 3109
to 3 106
Infrared (IR) 6106
to 7600
Visible 7600 to 3800
Ultra violet (UV) 3800 to 150
X–rays 150 to 0.1
Gamma rays 0.1 to 0.01
Cosmic rays 0.01 to zero
5. Atomic Spectrum
If the atom gains energy the electron passes from a lower energy level to a higher energy
level, energy is absorbed that means a specific wave length is absorbed. Consequently, a
dark line will appear in the spectrum. This dark line constitutes the absorption spectrum.
If the atom loses energy, the electron passes from higher to a lower energy level, energy is
released and a spectral line of specific wavelength is emitted. This line constitutes the
emission spectrum.
Hydrogen Atom
If an electric discharge is passed through hydrogen gas taken in a discharge tube under low
pressure, and the emitted radiation is analysed with the help of spectrograph, it is found to
consist of a series of sharp lines in the UV, visible and IR regions. This series of lines is
known as line or atomic spectrum of hydrogen. The lines in the visible region can be directly
seen on the photographic film.
Each line of the spectrum corresponds to a light of definite wavelength. The entire spectrum
consists of six series of lines each series, known after their discoverer as the Balmer,
Paschen, Lyman, Brackett, Pfund and Humphrey series. The wavelength of all these series
can be expressed by a single formula.



1
= R 







 2
2
2
1
1
1
n
n
Where,
 = wave number
 = wave length
R = Rydberg constant (109678 cm–1
)
n1 and n2 have integral values as follows

8. Series n1 n2 Main spectral lines
Lyman 1 2,3,4, etc Ultra – violet
Balmer 2 3,4,5 etc Visible
Paschen 3 4,5,6 etc Infra – red
Brackett 4 5,6,7 etc Infra – red
Pfund 5 6,7,8, etc Infra – red
[Note: All lines in the visible region are of Balmer series but reverse is not true. i.e., all
Balmer lines will not fall in visible region]
The pattern of lines in atomic spectrum is characteristic of hydrogen.
Types of Emission spectra
i) Continuous spectra: When white light from any source such as sun or bulb is
analysed by passing through a prism, it splits up into seven different wide bands of
colour from violet to red (like rainbow). These colour are so continuous that each of
them merges into the next. Hence the spectrum is called as continuous spectrum.
ii) Line spectra: When an electric discharge is passed through a gas at low pressure light
is emitted. If this light is resolved by a spectroscope, It is found that some isolated
coloured lines are obtained on a photographic plate separated from each other by dark
spaces. This spectrum is called line spectrum. Each line in the spectrum corresponds to
a particular wavelength. Each element gives its own characteristic spectrum.
6. Planck’s quantum theory
When a black body is heated, it emits thermal radiations of different wavelengths or
frequency. To explain these radiations, Max Planck put forward a theory known as Planck’s
quantum theory. The main points of quantum theory are
i) Substances radiate or absorb energy discontinuously in the form of small packets or
bundles of energy.
ii) The smallest packet of energy is called quantum. In case of light the quantum is known
as photon.
iii) The energy of a quantum is directly proportional to the frequency of the radiation . E  
(or) E = h were  is the frequency of radiation and h is Planck’s constant having the
value 6.626  10–27
erg – sec or 6.626  10–34
J–sec.
iv) A body can radiate or absorb energy in whole number multiples of a
quantum h, 2h,3h………..nh. where n is the positive integer.
Neils Bohr used this theory to explain the structure of atom.

9. 7. Bohr’s Atomic Model
Bohr developed a model for hydrogen atom and hydrogen like one–electron species
(hydrogenic species). He applied quantum theory in considering the energy of an electron
bound to the nucleus.
Important postulates
 An atom consists of a dense nucleus situated at the centre with the electron revolving
around it in circular orbits without emitting any energy. The force of attraction between
the nucleus and an electron is equal to the centrifugal force of the moving electron.
 Of the finite number of circular orbits around the nucleus, an electron can revolve only in
those orbits whose angular momentum (mvr) is an integral multiple of factor 
2
h
mvr =

2
nh
where, m = mass of the electron
v = velocity of the electron
n = orbit number in which electron is present
r = radius of the orbit
 As long as an electron is revolving in an orbit it neither loses nor gains energy. Hence
these orbits are called stationary states. Each stationary state is associated with a
definite amount of energy and it is also known as energy levels. The greater the
distance of the energy level from the nucleus, the more is the energy associated with it.
The different energy levels are numbered as 1,2,3,4, ( from nucleus onwards) or
K,L,M,N etc.
 Ordinarily an electron continues to move in a particular stationary state without losing
energy. Such a stable state of the atom is called as ground state or normal state.
 If energy is supplied to an electron, it may jump (excite) instantaneously from lower
energy (say 1) to higher energy level (say 2,3,4, etc.) by absorbing one or more quanta
of energy. This new state of electron is called as excited state. The quantum of energy
absorbed is equal to the difference in energies of the two concerned levels.
Since the excited state is less stable, atom will lose it’s energy and come back to the
ground state.
Energy absorbed or released in an electron jump, (E) is given by
E = E2 – E1 = h
where E2 and E1 are the energies of the electron in the first and second energy levels,
and  is the frequency of radiation absorbed or emitted.
[Note: If the energy supplied to hydrogen atom is less than 13.6 eV, it will accept or
absorb only those quanta which can take it to a certain higher energy level i.e.,

10. all those photons having energy less than or more than a particular energy level
will not be absorbed by hydrogen atom. But if energy supplied to hydrogen atom
is more than 13.6 eV then all photons are absorbed and excess energy appear
as kinetic energy of emitted photo electron].
Radius and Energy levels of hydrogen atom
Consider an electron of mass ‘m’ and charge ‘e’ revolving around a nucleus of charge Ze
(where, Z = atomic number and e is the charge of the proton) with a tangential
velocity v. r is the radius of the orbit in which electron is revolving.
By Coulomb’s Law, the electrostatic force of attraction between the moving electron
and nucleus is
Coulombic force = 2
2
r
KZe
K =
o


4
1
(where o is permittivity of free space)
K = 9 109
Nm2
C–2
In C.G.S. units, value of K = 1 dyne cm2
(esu)–2
The centrifugal force acting on the electron is
r
mv2
Since the electrostatic force balance the centrifugal force, for the stable electron orbit.
r
mv2
= 2
2
r
KZe
… (1)
(or) v2
=
mr
KZe 2
… (2)
According to Bohr’s postulate of angular momentum quantization, we have
mvr =

2
nh
v =
mr
2
nh

v2
= 2
2
2
2
2
4 r
m
h
n

… (3)
Equating (2) and (3)
2
2
2
2
2
2
4 r
m
h
n
mr
KZe


solving for r we get r = 2
2
2
2
mKZe
4π
h
n
where n = 1,2,3 - - - - - 

11. Hence only certain orbits whose radii are given by the above equation are available for the
electron. The greater the value of n, i.e., farther the energy level from the nucleus the
greater is the radius.
The radius of the smallest orbit (n=1) for hydrogen atom (Z=1) is ro.
ro =
K
me
h
n
2
2
2
2
4
=
 
    9
2
19
31
2
2
34
2
10
9
10
6
.
1
10
9
14
.
3
4
10
626
.
6
1












= 5.29 10–11
m = 0.529 Å
Radius of nth
orbit for an atom with atomic number Z is simply written as
rn = 0.529 
Z
n2
Å
Calculation of energy of an electron
The total energy, E of the electron is the sum of kinetic energy and potential energy.
Kinetic energy of the electron = ½ mv2
Potential energy =
r
e
K 2
Z

Total energy = 1/2 mv2
–
r
e
K 2
Z
… (4)
From equation (1) we know that
r
mv2
= 2
2
r
KZe
 ½ mv2
=
r
KZe
2
2
Substituting this in equation (4)
Total energy (E) =
r
2
KZe2
–
r
KZe2
=
r
2
KZe2

Substituting for r, gives us
E = 2
2
2
4
2
2
h
n
K
e
mZ
2π
where n = 1,2,3……….
This expression shows that only certain energies are allowed to the electron. Since this
energy expression consist of so many fundamental constant, we are giving you the following
simplified expressions.
E = –21.8 10–12
 2
2
n
Z
erg per atom
= –21.8 10–19
 2
2
n
Z
J per atom = –13.6  2
2
n
Z
eV per atom
(1eV = 3.83 10–23
kcal

12. 1eV = 1.602 10–12
erg
1eV = 1.602 10–19
J)
E = –313.6  2
2
n
Z
kcal / mole (1 cal = 4.18 J)
The energies are negative since the energy of the electron in the atom is less than the
energy of a free electron (i.e., the electron is at infinite distance from the nucleus) which is
taken as zero. The lowest energy level of the atom corresponds to n=1, and as the quantum
number increases, E becomes less negative.
When n = , E = 0, which corresponds to an ionized atom i.e., the electron and nucleus are
infinitely separated.
H  H+
+ e–
(ionisation).
Exercise 1: Find out the value of electrostatic potential energy of two electrons
separated by 3.0Å in vacuum./ Express your answer in joules and
electron volt.
Explanation for hydrogen spectrum by Bohr’s theory
According to the Bohr’s theory electron neither emits nor absorbs energy as long as it stays
in a particular orbit. However, when an atom is subjected to electric discharge or high
temperature, and electron in the atom may jump from the normal energy level, i.e., ground
state to some higher energy level i.e, exited state. Since the life time of the electron in
excited state is short, it returns to the ground state in one or more jumps.
During each jump, energy is emitted in the form of a photon of light of definite wavelength or
frequency. The frequency of the photon of light thus emitted depends upon the energy
difference of the two energy levels concerned (n1, n2) and is given by
h = E2 – E1 = 2
2
4
2
2
2
h
K
e
mZ










 2
1
2
2
1
1
n
n
 = 3
2
4
2
2
2
h
K
e
mZ









 2
2
2
1
1
1
n
n
The frequencies of the spectral lines calculated with the help of above equation are found to
be in good agreement with the experimental values. Thus, Bohr’s theory elegantly explains
the line spectrum of hydrogen and hydrogenic species.
Bohr had calculated Rydberg constant from the above equation.
 =

c
= 3
2
4
2
2
h
K
e
mZ
2








 2
2
2
1
1
1
n
n
c
h
K
e
mZ
2
1
3
2
4
2
2




 







 2
2
2
1
1
1
n
n

13. where
c
h
K
me
2
3
2
4
2

= 1.097  10–7
m–1
or 109678 cm–1
i.e., Rydberg constant (R)

λ
1
ν  = RZ2






 2
2
2
1 n
1
n
1
According to quantum mechanics the Rydberg constant is given by R =
C
h
8
e
3
2
0
4


Were  is reduced mass for the atom 0 is the permitivity of a vacuum. The reduced mass of
an atom with one electron is given by
 =
e
nuc
e
nuc
m
m
m
m


where mnuc = mass of nucleus
and me = mass of electron. For a nucleus with mass mnuc  
 = me
and R = 109673 cm–1
Illustration 1: Find out the longest wavelength of absorption line for hydrogen gas
containing atoms in ground state.
Solution: 







 2
2
2
1
2
n
1
n
1
RZ
1
for longest wavelength E should be smallest i.e. transition occurs from
n = 1 to n = 2
n = 7
n = 6
n = 5
n = 4
n = 3
n = 2
n = 1
i.e.

1
= 109673 cm–1
 12






 2
2
1
1
1


1
= 109673 
4
3
cm–1
  = 1
cm
109673
3
4


= 1.2157  10–5
cm = 121.6 nm

14. Exercise 2: The series limit for the Paschen series of hydrogen spectrum occurs at
8205.8Å. Calculate.
a) Ionization energy of hydrogen atom
b) Wave length of the photon that would remove the electron in the
ground state of the hydrogen atom.
Exercise 3: Calculate frequency of the spectral line when an electron from 5th
Bohr
orbit jumps to the second Bohr orbit in a hydrogen atom.
Exercise 4: Calculate the energy of an electron in 3rd
Bohr orbit.
Illustration 2 Calculate the energy in kJ per mole of electronic charge accelerated by a
potential of 1 volt.
Solution: Energy in joules = charge in coulombs  potential difference in volt
= 1.6 10–19
 6.02  1023
 1
= 9.632  104
J or 96.32 kJ
Exercise 5: What is highest frequency photon that can be emitted from hydrogen
atom? What is wavelength of this photon?
Exercise 6: Calculate the longest wavelength transition in the Paschen series of He+
.
Exercise 7: Calculate the ratio of the wavelength of first and the ultimate line of Balmer
series of Li2+
?
Calculation of velocity
We know that
mvr =

2
nh
; v =
mr
2
nh

By substituting for r we are getting
v =
nh
KZe
2 2

Where excepting n and z all are constants
v = 2.18 108
n
Z
cm/sec.
Further application of Bohr’s work was made, to other one electron species (Hydrogenic
ion) such as He+
and Li2+
. In each case of this kind, Bohr’s prediction of the spectrum was
correct.
Merits of Bohr’s theory
i) The experimental value of radii and energies in hydrogen atom are in good agreement
with that calculated on the basis of Bohr’s theory.
ii) Bohr’s concept of stationary state of electron explains the emission and absorption
spectra of hydrogen like atoms.

15. iii) The experimental values of the spectral lines of the hydrogen spectrum are in close
agreement with that calculated by Bohr’s theory.
Limitations of Bohr’s theory
i) It does not explain the spectra of atoms having more than one electron.
ii) Bohr’s atomic model failed to account for the effect of magnetic field (Zeeman effect) or
electric field (Stark effect) on the spectra of atoms or ions. It was observed that when the
source of a spectrum is placed in a strong magnetic or electric field, each spectral line
further splits into a number of lines. This observation could not be explained on the basis
of Bohr’s model.
iii) De Broglie suggested that electrons like light have dual character. It has particle and
wave character. Bohr treated the electron only as particle.
iv) Another objection to Bohr’s theory came from Heisenberg’s Uncertainty Principle.
According to this principle “It is impossible to determine simultaneously the exact
position and momentum of a small moving particle like an electron”. The postulate
of Bohr, that electrons revolve in well defined orbits around the nucleus with well
defined velocities is thus not tenable.
Exercise 8: Calculate the de Broglie wavelength of an electron that has been
accelerated from rest through a potential difference of 1 KV.
Illustration 3: A series of lines in the spectrum of atomic hydrogen lies at wavelengths
656.46, 482.7, 434.17, 410. 29 nm. What is the wave length of next line in
this series.
Solution: The given series of lines are in the visible region and thus appears to be
Balmer series
Therefore n1 = 2 and n2 = ? for next line
If  = 410.29 10–7
cm and n1 = 2
n2 may be calculated for the last line

1
= R 





 2
2
2
1
1
1
n
n
7
10
29
.
410
1


= 109673 





 2
2
2
1
2
1
n
n2 = 6
thus next line will be obtained during the jump of electron from 7th
to 2nd
shell i.e,

1
= R 





 2
2
7
1
2
1
= 109673 






49
1
4
1
 = 397.2 10–7
cm = 397.2 nm

16. Exercise 9: Calculate wavelength of photon emitted when an electron goes from n = 3
to n = 2 level of hydrogen atom.
Quantum Numbers
An atom contains large number of shells and subshells. These are distinguished from one
another on the basis of their size, shape and orientation (direction) in space. The
parameters are expressed in terms of different numbers called quantum numbers.
Quantum numbers may be defined as a set of four numbers with the help of which we can
get complete information about all the electrons in an atom. It tells us the address of the
electron i.e., location, energy, the type of orbital occupied and orientation
of that orbital.
i) Principal quantum number (n): It tells the main shell in which the electron resides and
the approximate distance of the electron from the nucleus. It also tells the maximum
number of electrons a shell can accommodate is 2n2
, where n is the principal quantum
number.
Shell K L M N
Principal quantum number (n) 1 2 3 4
Maximum number of electrons 2 8 18 32
ii) Azimuthal or angular momentum quantum number (l): This represents the number
of subshells present in the main shell. These subsidiary orbits within a shell will be
denoted as 1,2,3,4,… or s,p,d,f… This tells the shape of the subshells. The orbital
angular momentum of the electron is given as
 
1

l
l

2
h (or)  
1

l
l for a particular value of ‘n’ 







2
h
where . For a given
value of n values of possible l vary from 0 to n – 1.
iii) The magnetic quantum number (m): An electron due to its angular motion around
the nucleus generates an electric field. This electric field is expected to produce a
magnetic field. Under the influence of external magnetic field, the electrons of a subshell
can orient themselves in certain preferred regions of space around the nucleus called
orbitals. The magnetic quantum number determines the number of preferred orientations
of the electron present in a subshell. The values allowed depends on the value of l, the
angular momentum quantum number, m can assume all integral values between –l to
+l including zero. Thus m can be –1, 0, +1 for l = 1. Total values of m associated with a
particular value of l is given by 2l + 1.
iv) The spin quantum number (s): Just like earth not only revolves around the sun but
also spins about its own axis, an electron in an atom not only revolves around the
nucleus but also spins about its own axis. Since an electron can spin either in clockwise
direction or in anticlockwise direction, therefore, for any particular value of magnetic
quantum number, spin quantum number can have two values, i.e., +1/2 and –1/2 or
these are represented by two arrows pointing in the opposite directions, i.e.,  and .
When an electron goes to a vacant orbital, it can have a clockwise or anti clockwise spin
i.e., +1/2 or –1/2. This quantum number helps to explain the magnetic properties of the
substances.

17. Shapes and size of orbitals
An orbital is the region of space around the nucleus within which the probability of finding an
electron of given energy is maximum (90–95%). The shape of this region (electron cloud)
gives the shape of the orbital. It is basically determined by the azimuthal quantum number l,
while the orientation of orbital depends on the magnetic quantum number (m). Let us now
see the shapes of orbitals in the various subshells.
s–orbitals: These orbitals are
spherical and symmetrical about the
nucleus. The probability of finding the
electron is maximum near the nucleus
and keep on decreasing as the
distance from the nucleus increases.
There is vacant space between two
successive s–orbitals known as radial
node. But there is no radial node for
1s orbital since it is starting from the
nucleus.
nucleus
Z radial node
2s
1s
x
y
The size of the orbital depends upon the value of principal quantum number(n). Greater the
value of n, larger is the size of the orbital. Therefore, 2s–orbital is larger than 1s orbital but
both of them are non-directional and spherically symmetrical in shape.
p–orbitals (l =1): The probability of finding the p–electron is maximum in two lobes on the
opposite sides of the nucleus. This gives rise to a dumb–bell shape for the p–orbital. For p–
orbital l = 1. Hence, m = –1, 0, +1. Thus, p–orbital have three different orientations. These
are designated as px,py & pz depending upon whether the density of electron is maximum
along the x y and z axis respectively. As they are not spherically symmetrical, they have
directional character. The two lobes of p–orbitals are separated by a nodal plane, where the
probability of finding electron is zero.
Y
pz
py
px
Z
X
The three p-orbitals belonging to a particular energy shell have equal energies and are
called degenerate orbitals.
d–orbitals (l =2): For d–orbitals, l =2. Hence m=–2,–1,0,+1,+2. Thus there are 5 d orbitals.
They have relatively complex geometry. Out of the five orbitals, the three (dxy, dyz,dzx) project
in between the axis and the other two 2
z
d and 2
2
y
d 
x
lie along the axis.

18. Y
X
Z
dx y
2 2

dz2
Dough–nut shape or Baby
soother shape
Clover leaf shape
Z
Y
dxy
X X Y
Z
dyz
dxz
Rules for filling of electrons in various orbitals
The atom is built up by filling electrons in various orbitals according to the following rules.
Aufbau Principle: This principle states that the electrons are added one by one to the
various orbitals in order of their increasing energy starting with the orbital of lowest energy.
The increasing order of energy of various orbital is
1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,4f,5d,6p,5f,6d,7p……………………
How to remember such a big sequence? To make it simple we are giving you the method
to write the increasing order of the orbitals. Starting from the top, the direction of the arrows
gives the order of filling of orbitals.
1s
2s
3s
4s
5s
6s
7s
2p
3p
4p
5p
6p
3d
4d
5d
4f

19. Alternatively, the order of increasing energies of the various orbitals can be calculated on
the basis of (n+l) rule.
The energy of an orbital depends upon the sum of values of the principal quantum number
(n) and the azimuthal quantum number (l). This is called (n+ l) rule.
According to this rule,
“In neutral isolated atom, the lower the value of (n+ l) for an orbital, lower is its
energy. However, if the two different types of orbitals have the same value of
(n+ l), the orbitals with lower value of n has lower energy’’.
Illustration of (n + l) rule
Type of orbitals Value of n Values of l
Values of
(n+ l)
Relative energy
1s 1 0 1+0=1 Lowest energy
2s 2 0 2+0=2 Higher energy than 1s orbital
2p 2 1 2+1=3 2p orbital (n=2) have lower
energy than 3s orbital (n=3)
3s 3 0 3+1=3
Pauli’s Exclusion principle
According to this principle, an orbital can contain a maximum number of two electrons and
these two electrons must be of opposite spin.
Two electrons in an orbital can be represented by
 or 
Hund’s rule of maximum multiplicity
This rule deals with the filling of electrons in the equal energy (degenerate) orbitals of the
same sub shell (p,d and f). According to this rule,
“Electron pairing in p,d and f orbitals cannot occur untill each orbital of a given subshell
contains one electron each or is singly occupied”.
This is due to the fact that electrons being identical in charge, repel each other when
present in the same orbital. This repulsion can, however, be minimised if two electrons move
as far apart as possible by occupying different degenerate orbitals. All the electrons in a
degenerate set of orbitals will have same spin.
Illustration 4: Which hydrogen like ionic species has wavelength difference between the
first line of Balmer and first line of Lyman series equal to 59.3  10–9
m?
Neglect the reduced mass effect.
Solution: Wave number of first Balmer line of an species with atomic number Z is
given by








 2
2
2
2
1
1
1
RZ
v

20. 36
RZ
5
v
2

 ; Similarly wave number of v of first Lyman line is given by
v = RZ2






 2
2
2
1
1
1
= 2
RZ
4
3
;





1
v
and
1
v
  –  = 2
2
RZ
3
4
RZ
5
36
 = 






3
4
5
36
RZ
1
2
= 2
RZ
15
88
Z2
= 7
9
10
097
.
1
15
10
3
.
59
88



 
= 9 or Z = 3  ionic species is Li2+
Electronic configuration of elements
Electronic configuration is the distribution of electrons into different shells, subshells and
orbitals of an atom .
Keeping in view the above mentioned rules, electronic configuration of any orbital can be
simply represented by the notation.]
n l
x
Number of electrons in the
subshell
Symbol of subshell or
orbitals (s,p,d,f)
Principal quantum number
Alternatively
Orbital can be represented by a box and an electron with its direction of spin by arrow. To
write the electronic configuration, just we need to know (i) the atomic number (ii) the order in
which orbitals are to be filled (iii) maximum number of electrons in a shell, sub–shell or
orbital.
a) Each orbital can accommodate two electrons
b) The number of electrons to be accomodated in a subshell is 2  number of
degenerate orbitals.
Subshell Maximum number of electrons
s 2
p 6
d 10
f 14
c) The maximum number of electron in each shell (K,L,M,N…) is given by 2n2
.
where n is the principal quantum number.
d) The maximum number of orbitals in a shell is given by n2
where n is the principal
quantum number.
e) The number of nodal planes associated with an orbital is given by l -1.

21. Importance of knowing the electronic configuration
The chemical properties of an element are dependent on the relative arrangement
of its electrons.
Illustration 5: Write the electronic configuration of nitrogen (atomic number= 7)
Solution:
1s2
2s2
2p3
    
Exceptional Configurations
Stability of half filled and completely filled orbitals
Cu has 29 electrons. Its expected electronic configuration is
1s2
2s2
2p6
3s2
3p6
4s2
3d9
But a shift of one electron from lower energy 4s orbital to higher energy 3d orbital will make
the distribution of electron symmetrical and hence will impart more stability.
Thus the electronic configuration of Cu is
1s2
2s2
2p6
3s2
3p6
4s1
3d10
Fully filled and half filled orbitals are more stable.
Illustration 6: We know that fully filled and half filled orbitals are more stable. Can you
write the electronic configuration of Cr(Z = 24)?.
Solution: Cr (Z = 24)
1s2
, 2s2
,2p6
,3s2
,3p6
,4s1
,3d5
.
Since half filled orbital is more stable one 4s electron is shifted to 3d
orbital.
Exercise 10: 1st
I.P. of nitrogen is higher than oxygen. Explain.
8. Dual Character
(Particle and Wave Character of Matter and Radiation)
In case of light some phenomenon like diffraction and interference can be explained on the
basis of its wave character. However, the certain other phenomenon such as black body
radiation and photoelectric effect can be explained only on the basis of its particle nature.
Thus, light is said to have a dual character. Such studies on light were made by Einstein in
1905.
Louis de Broglie, in 1924 extended the idea of photons to material particles such as
electron and he proposed that matter also has a dual character-as wave and
as particle.

22. Derivation of de-Broglie equation
The wavelength of the wave associated with any material particle was calculated by analogy
with photon.
In case of photon, if it is assumed to have wave character, its energy is given by
E = h ------------------ (i) (according to the Planck’s quantum theory)
where  is the frequency of the wave and ‘h’ is Planck’s constant
If the photon is supposed to have particle character, its energy is given by
E = mc2
------------------ (ii) (according to Einstein’s equation)
where ‘m’ is the mass of photon, ‘c’ is the velocity of light.
By equating (i) and (ii)
h = mc2
But  = c/
h

c
= mc2
(or)  = h /mc
The above equation is applicable to material particle if the mass and velocity of photon is
replaced by the mass and velocity of material particle. Thus for any material particle like
electron.
 = h/mv (or) =
p
h
where mv = p is the momentum of the particle.
9. Derivation of Angular Momentum from de Broglie
Equation
According to Bohr’s model, the electron revolves around the nucleus in circular orbits.
According to de Broglie concept, the electron is not only a particle but has a wave character
also.
If the wave is completely in phase, the circumference of the orbit
must be equal to an integral multiple of wave length ()
Therefore 2r = n
where ‘n’ is an integer and ‘r’ is the radius of the orbit
But  = h/mv
 2r = nh /mv
(or) mvr = n h/2

23. which is Bohr’s postulate of angular momentum, where ‘n’ is the principal
quantum number.
“Thus, the number of waves an electron makes in a particular Bohr orbit in one complete
revolution is equal to the principal quantum number of the orbit”.
Alternatively
Number of waves ‘n’ =

r
2
=
mv
h
r
2
=
h
mvr
2
where v and r are the velocity of electron and radius of that particular Bohr orbit in which
number of waves are to be calculated, respectively.
The electron is revolving around the nucleus in a circular orbit. How many revolutions
it can make in one second
Let the velocity of electron be v m/sec. The distance it has to travel for one revolution 2r,
(i.e., the circumference of the circle).
Thus, the number of revolutions per second is =
r
v

2
Common unit of energy is electron volt which is amount of energy given when an electron is
accelerated by a potential of exactly 1 volt. This energy equals the product of voltage and
charge. Since in SI units coulombs  volts = Joules, 1 eV numerically equals the electronic
charge except that joule replaces coulombs.
Illustration 7: Two particles A and B are in motion. If the wavelength associated with
particle A is 5 10–8
m, calculate the wavelength associated with particle B
if its momentum is half of A.
Solution: According to de Broglie equation
A =
A
p
h
and B =
B
p
h
B
A


=
A
B
p
p
But pB = ½ pA (given)
B
A


=
A
A
p
p
2
/
1
= ½
B = 2A = 2  510–8
m = 10–7
m
Illustration 8: Calculate the de Broglie wavelength of a ball of mass 0.1kg moving with a
speed of 60ms–1
.
Solution:
mv
h

 =
60
1
.
0
10
6
.
6 34

 
 = 1.1  10–34
m.

24. This is apparent that this wavelength is too small for ordinary observation.
Although the de Broglie equation is applicable to all material objects but it
has significance only in case of microscopic particles.
Since, we come across macroscopic objects in our everyday life, de
Broglie relationship has no significance in everyday life.
[Distinction between the wave- particle nature of a photon and the particle- wave
nature of a sub atomic particle]
Photon Sub Atomic Particle
1. Energy = h
Energy =
2
1
mv2
2. Wavelength =

c
Wavelength =
mv
h
[Note: We should never interchange any of the above]
10. Heisenberg’s Uncertainty Principle
All moving objects that we see around us e.g., a car, a ball thrown in the air etc., move along
definite paths. Hence their position and velocity can be measured accurately at any instant
of time. Is it possible for subatomic particle also?
As a consequence of dual nature of matter, Heisenberg, in 1927 gave a principle about the
uncertainties in simultaneous measurement of position and momentum (mass  velocity) of
small particles.
This principle states
“It is impossible to measure simultaneously the position and momentum of a small
microscopic moving particle with absolute accuracy or certainty” i.e., if an attempt is made to
measure any one of these two quantities with higher accuracy, the other becomes less
accurate.
The product of the uncertainty in position (x) and the uncertainty in the momentum (p =
m.v where m is the mass of the particle and v is the uncertainty in velocity) is equal to or
greater than h/4 where h is the Planck’s constant.
Thus, the mathematical expression for the Heisenberg’s uncertainty principle is simply
written as
x . p  h/4
Explanation of Heisenberg’s uncertainty principle
Suppose we attempt to measure both the position and momentum of an electron, to pinpoint
the position of the electron we have to use light so that the photon of light strikes the
electron and the reflected photon is seen in the microscope. As a result of the hitting, the
position as well as the velocity of the electron are disturbed. The accuracy with which the

25. position of the particle can be measured depends upon the wavelength of the light used.
The uncertainty in position is . The shorter the wavelength, the greater is the accuracy.
But shorter wavelength means higher frequency and hence higher energy. This high energy
photon on striking the electron changes its speed as well as direction. But this is not true for
macroscopic moving particle. Hence Heisenberg’s uncertainty principle is not applicable to
macroscopic particles.
Illustration 9: Why electron cannot exist inside the nucleus according to Heisenberg’s
uncertainty principle?
Solution: Diameter of the atomic nucleus is of the order of 10–15
m
The maximum uncertainty in the position of electron is 10–15
m.
Mass of electron = 9.1 10–31
kg.
x. p =

4
h
x  (m.v) = h/4
v =
m
.
x
1
4
h



=
7
22
4
10
63
.
6 34

 
 31
15
10
1
.
9
10
1




v = 5.80  1010
ms–1
This value is much higher than the velocity of light and
hence not possible.
11. Quantum Mechanical Model of atom
The atomic model which is based on the particle and wave nature of the electron is known
as wave or quantum mechanical model of the atom. This was developed by Ervin
Schrodinger in 1926. This model describes the electron as a three dimensioinal wave in the
electronic field of positively charged nucleus. Schrodinger derived an equation which
describes wave motion of an electron. The differential equation is
0
)
V
E
(
h
m
8
dz
d
dy
d
dx
d
2
2
2
2
2
2
2
2










where x, y, z are certain coordinates of the electron, m = mass of the electron E = total
energy of the electron. V = potential energy of the electron; h = planck’s constant and z
(psi) = wave function of the electron.
Significance of  The wave function may be regarded as the amplitude function expressed
in terms of coordinates x, y and z. The wave function may have positive or negative values
depending upon the value of coordinates. The main aim of Schrodinger equation is to give
solution for probability approach. When the equation is solved, it is observed that for some
regions of space the value of  is negative. But the probability must be always positive and
cannot be negative, it is thus, proper to use 2
in favour of .
Significance of 2
: 2
is a probability factor. It describes the probability of finding an
electron within a small space. The space in which there is maximum probability of finding an

26. electron is termed as orbital. The important point of the solution of the wave equation is that
it provides a set of numbers called quantum numbers which describe energies of the
electron in atoms, information about the shapes and orientations of the most probable
distribution of electrons around nucleus.
12. Photo Electric Effect
Sir J.J. Thomson, observed that when a light of certain frequency strikes the surface of a
metal, electrons are ejected from the metal. This phenomenon is known as photoelectric
effect and the ejected electrons are called photoelectrons.
A few metals, which are having low ionisation energy like Cesium, show this effect under the
action of visible light but many more show it under the action of more energetic ultraviolet
light.
V
A
–
+
electrons
Evacuated quartz tube
Light
An evacuated tube contains two electrodes connected to a source of variable voltage, with
the metal plate whose surface is irradiated as the anode. Some of the photoelectrons that
emerge from this surface have enough energy to reach the cathode despite its negative
polarity, and they constitute the measured current. The slower photoelectrons are repelled
before they get to the cathode. When the voltage is increased to a certain value Vo, of the
order of several volts, no more photoelectrons arrive, as indicated by the current dropping to
zero. This extinction voltage (or also referred as stopping potential) corresponds to the
maximum photoelectron kinetic energy i.e., eVo = ½ mv2
The experimental findings are summarised as below:
i) Electrons come out as soon as the light (of sufficient energy) strikes
the metal surface .
ii) The light of any frequency will not be able to cause ejection of electrons from a metal
surface. There is a minimum frequency, called the threshold (or critical) frequency, which
can just cause the ejection. This frequency varies with the nature of the metal. The
higher the frequency of the light, the more energy the photoelectrons have. Blue light
results in faster electrons than red light.
iii) Photoelectric current is increased with increase in intensity of light of same frequency, if
emission is permitted i.e., a bright light yields more photoelectrons than a dim one of the
same frequency, but the electron energies remain the same.

27. Light must have stream of energy particles or quanta of energy (h). Suppose, the threshold
frequency of light required to eject electrons from a metal is o, when a photon of light of
this frequency strikes a metal it imparts its entire energy (ho) to the electron.
E = ho E > ho
K.Emax = h – ho
K. E = 0
Metal
“This energy enables the electron to break away from the atom by overcoming the attractive
influence of the nucleus”. Thus each photon can eject one electron. If the frequency of light
is less than o there is no ejection of electron. If the frequency of light is higher than o (let it
be ), the photon of this light having higher energy (h), will impart some energy to the
electron that is needed to remove it away from the atom. The excess energy would give a
certain velocity (i.e, kinetic energy) to the electron.
h = ho + K.E
h = ho + ½ mv2
½ mv2
= h–ho
where,  = frequency of the incident light
o = threshold frequency
ho is the threshold energy (or) the work function denoted by  = ho (minimum energy of the
photon to liberate electron). It is constant for particular metal and is also equal to the
ionization potential of gaseous atoms.
The kinetic energy of the photoelectrons increases linearly with the frequency of incident
light. Thus, if the energy of the ejected electrons is plotted as a function of frequency, it
result in a straight line whose slope is equal to Planck’s constant ‘h’ and whose
intercept is ho.

K.E
of
Photoelectrons
Illustration 10: A photon of wavelength 5000 A strikes a metal surface, the work function of the
metal being 2.20 eV. Calculate (i) the energy of the photon in eV (ii) the kinetic
energy of the emitted photo electron and (iii) the velocity of the photo electron.

28. Solution: i) Energy of the photon
E = h =

hc
=
  
m
10
5
ms
10
3
Js
10
6
.
6
7
1
8
34






= 3.96  10–19
J
1 eV = 1.6  10–19
J
Therefore E =
ev
/
J
10
6
.
1
J
10
96
.
3
19
19




= 2.475 eV
ii) Kinetic energy of the emitted photo electron
Work function = 2.20 eV
Therefore, KE = 2.475 – 2.20 = 0.275 eV = 4.4  10–20
J
iii) Velocity of the photo electron
KE = 2
mv
2
1
= 4.4  10–20
J
Therefore, velocity (v) = 31
20
10
1
.
9
10
4
.
4
2





= 3.11  105
ms–1

29. 13. Solution to Exercises
Exercise 1: E =
r
KZe 2
K =
0
4
1

= 9  109
Nm2
C–2
= 10
2
19
9
10
3
)
10
6
.
1
(
1
10
9







= 7.68  10–19
J
 1 eV = 1.6  10–19
J
 E = 19
19
10
6
.
1
10
68
.
7




= 4.8 eV
Exercise 2: a) Energy corresponding to 8205.8 Ao
= 10
8
34
10
8
.
8205
10
3
10
626
.
6






= 2.422 × 10-19
J = 1.572 eV










 2
2
2
1
2
H
1
n
1
n
1
Z
E
E
Δ
1.512 eV = E1H × (1)2
× 






 2
2
1
3
1
1.512 eV =
9
E H
1
E1 H = 13.608 eV
 Ionisation energy of hydrogen atom = 13.6 eV
b) 19
8
34
10
602
.
1
6
.
13
10
3
10
626
.
6
E
hc










= 916 Ao
Exercise 3:

1
= 





 2
2
2
1
n
1
n
1
R = 109673 





 2
2
5
1
2
1
= 2.304 106
m–1
 v =

C
= 2.304  106
m–1
 2.998 108
m/s
= 6.906  1014
Hz
Exercise 4: En = 2
n
6
.
13
 eV
= 2
3
6
.
13
 = – 1.51 eV
= – 2.42  10–19
J
Exercise 5: Highest frequency photon is emitted when electron comes from infinity to
1st
energy level.

30. E = 2
2
1
Z
6
.
13
 = – 13.6eV
or 13.6  1.6  10–19
Joule = 2.176  10–18
Joule
E = h
 v =
h
E
=
JS
10
626
.
6
J
10
176
.
2
34
18




= 0.328  1016
Hz
v =

C
  = 16
8
10
328
.
0
10
3


= 9.146  10–8
m
Exercise 6:











 2
2
2
1
2
H
n
1
n
1
Z
R
For He; Z = 2; For Paschen series n1 = 3
For longest wavelength n2 = 4
  









 2
2
2
4
1
3
1
2
109678
1
= 109678×4× 






16
1
9
1
= 109678 ×4×
144
7
 = 4689 Ao
Exercise 7: wave number of first line of Balmer, 1
 =
4
5
36
9
5
3
1
2
1
2
2
2 R
R
RZ 









wave length of first line of Balmer =
R
5
4
wave number of ultimate line of Balmer, 2
 = 







1
2
1
2
2
RZ =
4
9R
wave length of ultimate line of Balmer =
R
9
4
Ratio =
5
9
Exercise 8: Energy in Joules = Charge on the electron in coloumb  pot. diff. in volts.
= 1.609 10–19
1000 = 1.609 10–16
J
Kinetic energy (1/2 mv2
) = 1.609 10–16
J
2
1
9.1 10–31
v2
= 1.609 10–16
v2
= 3.5361014
v = 1.88 107
ms–1
h = 6.626 10–34
J sec
 =
mv
h
= 7
31
34
10
88
.
1
10
1
.
9
10
626
.
6






= 3.87 10–11
m.

31. Exercise 9: 







 2
2
2
1
2
n
1
n
1
RZ
1
 








 2
2
2
3
1
2
1
1
109673
1
= 109673 
36
5
  =
5
109673
36

= 6.56  10–7
m or 656 nm
Exercise 10: Due to presence of half filled orbital in nitrogen which impart it extra
stability.

32. 14. Solved Problems
14.1 Subjective
Problem 1: Which electronic transition in Balmer series of hydrogen atom has same
frequency as that of n = 6 to n = 4 transition in He+
. [Neglect reduced mass
effect].
Solution: 








2
2
2
He
6
1
4
1
RZ
v = 







16
36
16
36
R
4 =
36
R
5
H
v = R  12






 2
2
n
1
2
1
 H
He
v
v 

36
R
5
= 2
n
R
4
R

On solving above equation
n2
= 9
 n = 3
Or corresponding transition from 3  2 in Balmer series of hydrogen atom
has same frequency as that of 6  4 transition in He+
.
Problem 2: Calculate ionisation potential in volts of (a) He+
and (b) Li2+
Solution: I.E. = 2
2
n
Z
6
.
13
= 2
2
1
Z
6
.
13 
[Z =2 for He+
] = 13.6  4 = 54.4 eV
Similarly for Li2+
= 2
2
1
3
6
.
13 
= 13.6  9 = 122.4 eV
Problem 3: What fraction of the volume of an atom of radius 10–8
cm is occupied by its
nucleus if nuclear radius is 10–12
cm?
Solution: Assuming atom to be spherical having definite boundary its volume can be
given by 3
r
3
4
 (where r is atomic radius). Similarly volume of nucleus can be
given by 3
r
3
4

 where r is radius of nucleus.
3
3
r
3
4
r
3
4
atom
of
Volume
nucleus
of
Volume








=
 
3
8
3
12
3
3
)
10
(
10
r
r




= 24
36
10
10


= 10–12

33. Problem4: Calculate the ratio of K.E and P.E of an electron in a orbit?
Solution: K.E. =
r
Ze
2
2
P.E. =
r
Ze2

 P.E. = –2K.E

2
1
P.E
K.E


Problem 5: How many spectral lines are emitted by atomic hydrogen excited to
nth
energy level?
Solution:
1 1+2=3
n = 1
n = 2
n = 3
n = 4
n = 5
n = 6
1+2+3= 6
1
2
3
1
2
3
4
3
1
5 6
Thus the number of lines emitted from nth energy level
= 1+2+3+ …………. n–1 =  (n–1)
n =
 
2
1

n
n
 (n–1)=
  
2
1
1
1 

 n
n
=
  
2
1 n
n 
Number of spectral lines that appear in hydrogen spectrum when an electron
de excites from nth
energy level =
 
2
1
n
n 
Problem 6: Calculate (a) the de Broglie wavelength of an electron moving with a velocity of
5.0  105
ms–1
and (b) relative de Broglie wavelength of an atom of hydrogen
and atom of oxygen moving with the same velocity (h = 6.63  10–34
kg m2
s–1
)
Solution: a)  =
mv
h
=
  
1
5
31
2
2
34
ms
10
0
.
5
kg
10
11
.
9
s
kgm
10
63
.
6







Wavelength  = 1.46  10–9
m
b) An atom of oxygen has approximately 16 times the mass of an atom of
hydrogen. In the formula
mv
h

 , h is constant while the conditions of
problem make v, also constant. This means that  and m are variables
and  veries inversely with m. Therefore,  for the hydrogen atom would
be 16 times greater than  for oxygen atom.

34. Problem 7: In photoelectric effect, an absorbed quantum of light results in the ejection of
an electron from the observor. The kinetic energy of the electron is equal to the
energy of the absorbed photon minus the energy of the longest wavelength
that causes the effect calculate the kinetic energy of an electron produced in
cesium by 400 nm light. The critical (maximum) wavelength for the
photoelectric effect is cesium in 660 nm.
Solution: Kinetic energy of electron = h – hcritical =

hc
-
critical
hc

KE =
nm
660
eV
nm
1240
nm
400
eV
nm
1240
 = 1.22 eV
[(Since 1240 nm = 1eV)
 = 1200  10–9
m
h = 6.626  10–34
 E = 9
8
34
0
`
1240
10
3
10
626
.
6






= 1.6  10–19
J or 1 eV]
Problem 8: It has been found that gaseous Iodine molecules dissociate into seperated
atoms after absorption of light at wavelengths less than 4995Å. If each
quantum is absorbed by one molecule of I2, what is the minimum input in
kcal/mole, needed to dissociate I2 by this photo chemical process.
Solution: E (per mole) = NAh
E = Na
  
m
10
4995
ms
10
3
Js
10
626
.
6
mol
10
023
.
6
hc
10
1
8
34
1
23











= 239.5 kJ/mol 





kJ
184
.
4
kcal
1
= 57.1 kcal/mole
Problem 9: What is the wavelength associated with 150 eV electron
Solution:  =
.
E
.
K
m
2
h


=
J
10
6
.
1
150
kg
10
1
.
9
2
JS
10
626
.
6
19
31
34









=
50
34
10
4368
10
626
.
6




= 10–10
m = 1Å
Problem 10: The energy of electron in the second and third Bohr orbit of the hydrogen atom
is –5.42  10–12
erg and –2.41  10–12
erg, respectively. Calculate the
wavelengths of emitted radiation when the electron drops from third to second
orbit.
Solution: E3 – E2 = h =

hc
– 2.41  10–12
– (– 5.42  10–12
) =



  10
27
10
3
10
626
.
6
  = 12
10
27
10
01
.
3
10
3
10
626
.
6






= 6.604  10–5
cm = 6.604  10–5
108
= 6604 Ă

35. Problem 11: Show that frequencies of emitted photons are additive but their wavelengths
are not.
Solution: n = 3
n = 2
n = 1
E2  3
E1  2
X
E1  3
E
 E1  3 = E1  2 + E2  3
h1  3 = h1  2 + h2  3
1  3 = 1  2 + 2  3
i.e. frequencies like energies are additive, on the other hand
 E1  3 = E1  2 + E2  3

3
2
2
1
3
1
hc
hc
hc









3
2
2
1
3
1
1
1
1









3
2
2
1
2
1
3
2
3
1
1




 







3
2
2
1
3
2
2
1
3
1












i.e. wavelengths are not additive.
Problem 12: The second ionization potential of Be is 17.98 eV. If the electron in Be+
is
assumed to move in a spherical orbit with a central field of effective nuclear
charge (Zeff) consisting of the nucleus and other electrons, by how many units
of charge is the nucleus shielded by other electrons?. (The energy of electron
in first Bohr orbit of H is –13.6 eV). If the extent of shielding by the K electrons
of Li is same as you have calculated above, find the ionization potential of Li.
Solution: Ionization Energy = 2
2
6
.
13
n
Zeff
= 17.98
 2
eff
Z =
6
.
13
2
98
.
17 2

= 5.28
Zeff = 2.3
Shielding effect = Z–Zeff = 4–2.3 = 1.7
Zeff for Lithium = 3–1.7 = 1.3
 IE of lithium =
 
2
2
2
3
.
1
6
.
13 
= 5.74 eV
Problem 13: Wavelength of the K characteristic X–rays of iron and potassium are
1.93110–8
and 3.737  10–8
cm respectively. What is the atomic number and

36. name of the element for which characteristic K wave length is
2.289 10–8
cm?
At. No of K = 19, and Fe = 26
Velocity of light = 3 1010
cm / sec
According to Mosly’s law, the frequency of emitted X – ray   Z2
Solution: Frequency  =

c
 Z2
=

c
K(Z)2
2
2
1
1
2
Z
Z











… (1)
2
3
2
2
3
Z
Z











… (2)
1 (Fe) = 1.931  10–8
, Z1 = 26
3(K) = 3.737 10–8
, Z3 = 19
2 (u) = 2.289  10–8
, Z2 = ?
from eqn (1)
2
2
8
8
Z
26
10
931
.
1
10
289
.
2













On solving we get Z2 = 23.88 = 24
From equation (2)
8
8
10
289
.
2
10
737
.
3




=
2
2
19
Z








Z2 = 24.27 = 24
Thus, the atomic number of that element is 24 and it is chromium.
Problem 14: O2 undergoes photochemical dissociation into one normal oxygen and one
excited oxygen atom, 1.967 eV more energetic than normal. The dissociation of
O2 into two normal atoms of oxygen atoms requires 498 KJ mole–1
. What is the
maximum wavelength effective for photochemical dissociation of O2?.
Solution: O2  ON + excited
O
O2  ON + ON
E = 498 103
J / mole
= J
23
3
10
023
.
6
10
498


per molecule = 8.268 10–19
J
Energy required for excitation = 1.967 eV = 3.146 10–19
J
Total energy required for photochemical dissociation of O2
= 8.268 10–19
+ 3.146 10–19
= 11.414  10–19
J

hc
= 11.414 10–19
J
 = 19
8
34
10
414
.
11
10
3
10
626
.
6






= 1.7415 10–7
m = 1741.5 Å

37. Problem 15: Compare the wavelengths for the first three lines in the Balmer series with
those which arise from similar transition in Be3+
ion. (Neglect reduced mass
effect).
Solution: 







 2
2
2
H
n
1
2
1
1
R
v








 2
2
2
Be
n
1
2
1
4
R
v

Be
H
H
Be
v
v


 = 16
So we can conclude that all transitions in Be3+
will occur at wavelengths
16
1
times the hydrogen wavelengths.
14.2 Objective
Problem 1: For a p-electron, orbital angular moment is
(A) 
2 (B) 
(C) 
6 (D) 
2
Solution: Orbital angular momentum L = 

 )
1
(  where


2
h

 L for p electron = 
 2
)
1
1
(
1 

 (A)
Problem 2: For which of the following species, Bohr theory doesn’t apply
(A) H (B) He+
(C) Li2+
(D) Na+
Solution: Bohr theory is not applicable to multi electron species
 (D)
Problem 3: If the radius of 2nd
Bohr orbit of hydrogen atom is r2. The radius of third Bohr
orbit will be
(A) 2
r
9
4
(B) 4r2
(C) 2
r
4
9
(D) 9r2
Solution: 2
2
2
2
mZe
4
h
n
r


 2
2
3
2
3
2
r
r

 r3
=
2
r
4
9
 (C)

38. Problem 4: Number of waves made by a Bohr electron in one complete revolution in 3rd
orbit is
(A) 2 (B) 3
(C) 4 (D) 1
Solution: Circumference of 3rd
orbit = 2r3
According to Bohr angular momentum of electron in 3rd
orbit is
mvr3 =

2
h
3
or
3
r
2
mv
h 3


by De-Broglie equation
 =
mv
h
  =
3
r
2 3

 2r3 = 3
i.e. circumference of 3rd
orbit is three times the wavelength of electron or
number of waves made by Bohr electron in one complete revolution in 3rd
orbit is three.
(B)
Problem 5: The degeneracy of the level of hydrogen atom that has energy
16
RH
 is
(A) 16 (B) 4
(C) 2 (D) 1
Solution: En = 2
H
n
R


16
R
n
R H
2
H



i.e. for 4th
sub-shell
- 1 0 +1 - 2 –1 +1
0 +2 – 3 – 2 – 1 0 +1 +2 +3
3
2
1
three p five d Seven f
n = 4
l = 0

m = 0
one s
i.e. 1 + 3 + 5 + 7 = 16
 degeneracy is 16
Problem 6: An electron is moving with a kinetic energy of 4.55 10–25
J. What will be
de Broglie wave length for this electron?
(A) 5.28 10–7
m (B) 7.28 10–7
m
(C) 2 10–10
m (D) 3 10–5
m
Solution: KE =
2
1
mv2
= 4.55  10–25
V2
= 6
31
25
10
1
10
1
.
9
10
55
.
4
2








39. V = 103
m/s
De Broglie wave length =
mv
h
= 3
31
34
10
10
1
.
9
10
626
.
6





= 7.28 10–7
m
 (B)
Problem 7: Suppose 10–17
J of energy is needed by the interior of human eye to see an
object. How many photons of green light (=550 nm) are needed to generate
this minimum amount of energy?
(A) 14 (B) 28
(C) 39 (D) 42
Solution: Let the number of photons required = n
n 17
10


hc
n =
hc


17
10
= 8
34
9
17
10
3
10
626
.
6
10
550
10








= 27.6 = 28 photons
 (B)
Problem 8: Photoelectric emission is observed from a surface for frequencies 1 and 2 of
the incident radiation (1>2). If the maximum kinetic energies of the
photoelectrons in two cases are in ratio 1:K then the threshold frequency 0 is
given by
(A)
1
K
1
2




(B)
1
K
2
1
K




(C)
1
K
K 1
2




(D)
K
1
2 


Solution: KE1 = h1 – ho
KE2 = h2–ho
It is given that
2
1
KE
KE
=
K
1
K
1
h
h
h
h
o
2
o
1







   
1
K
K o
2
1 





o =
1
K
K 2
1




 (B)
Problem 9: The velocity of electron in the ground state hydrogen atom is 2.18  106
ms–1
.
Its velocity in the second orbit would be
(A) 1.09 106
ms–1
(B) 4.38 106
ms–1
(C) 5.5 105
ms–1
(D) 8.76 106
ms–1-
Solution: We know that velocity of electron in nth Bohr’s orbit is given by
 = 2.18  106
n
Z
m/s

40. for H, Z = 1
 v1 =
1
10
18
.
2 6

m/s
 v2 =
2
10
18
.
2 6

m/s = 1.09 106
m/s
 (A)
Problem 10: The ionization energy of the ground state hydrogen atom is 2.1810–18
J. The
energy of an electron in its second orbit would be
(A)–1.09 10–18
J (B) –2.18 10–18
J
(C) –4.36 10–18
J (D) –5.45 10–19
J
Solution: Energy of electron in first Bohr’s orbit of H–atom
E = J
n2
18
10
18
.
2 


( ionization energy of H = 2.18 10–18
J)
E2 = 2
18
2
10
18
.
2 


J = –5.45 10–19
J
 (D)
Problem 11: Magnetic moments of V (Z = 23), Cr (Z = 24) and Mn (Z = 25) are x, y, z. Hence
(A) z  y  x (B) x = y = z
(C) x  z  y (D) x  y  z
Solution: Magnetic moments = .
M
.
B
)
2
n
(
n  where n is the number of unpaired
electron
V ( Z = 23) (Ar) 3d3
4s2
n = 3, 15 Bm = x
Cr (Z = 24) (Ar) 3d5
4s n = 6, 48 Bm = y
Mn (Z = 25) (Ar) 3d5
4s2
n = 5, 35 Bm = z
 C
Problem 12: The speed of a photon is one hundredth of the speed of light in vacuum. What
is the de Broglie wavelength? Assume that one mole of protons has a mass
equal to one gram. h = 6.626  10–27
erg sec
(A) 3.31  10–3
Å (B) 1.33  10–3
Å
(C) 3.13  10–2
Å (D) 1.31 10–2
Å
Solution: m = g
10
023
.
6
1
23

 =
mv
h
= 1
8
27
sec
cm
10
3
1
10
626
.
6





 6.023  1023
= 1.33  10–11
cm
 B
Problem 13: The wave number of first line of Balmer series of hydrogen atom is 15200 cm–1
.
What is the wave number of first line of Balmer series of Li2+
ion.
(A) 15200 cm–1
(B) 6080 cm–1
(C) 76000 cm–1
(D) 1,36800 cm–1

41. Solution: For He+
v
v  for H Z2
= 15200  9 = 1, 36,800
 D
Problem 14: The speed of the electron in the 1st orbit of the hydrogen atom in the ground
state is [C is the velocity of light]
(A)
1.37
C
(B)
1370
C
(C)
13.7
C
(D)
137
C
Solution: V =
mr
2
h

= 2.189  108
cm sec–1
C = 3  1010
cm,
v
C
= 137
 D
Problem 15: The quantum number not obtained from the Schrodinger’s wave equation is
(A) n (B) l
(C) m (D) s
Solution: n, l and m quantum numbers can be obtained from Schrodinger equation. S
is obtained from spectral evidence.
 D
Problem 16: Which of the following sets of quantum numbers is not allowed
(A) n = 3, l = 1, m = +2 (B) n = 3, l = 1, m = +1
(C) n = 3, l = 0, m = 0 (D) n = 3, l = 2, m =  2
Solution: If n = 3
l = 0,1,2
for l = 0, m = 0
l = 1, m = –1, 0, +1
l = 2 m = –2, –1, 0, +1, +2
If l = 1, the value of m can not be 2
 (A)
Problem 17: Assuming that a 25 watt bulb emits monochromatic yellow light of wave length
0.57 . The rate of emission of quanta per sec. will be
(A) 5.89  1015
sec–1
(B) 7.28  1017
sec–1
(C) 5  1010
sec–1
(D) 7.18  1019
sec–1
Solution: Let n quanta are evolved per sec.







hc
n = 25 J sec–1
n 6
8
34
10
57
.
0
10
3
10
626
.
6






= 25
n = 7.18  1019
sec–1
 (D)

42. Problem 18: How many chlorine atoms can you ionize in the process Cl  
Cl + e, by the
energy liberated from the following process.
Cl + e–
 
Cl for 6  1023
atoms
Given electron affinity of Cl = 3.61 eV, and I P of Cl = 17.422 eV
(A) 1.24  1023
atoms (B) 9.82  1020
atoms
(C) 2.02  1015
atoms (D) none of these
Solution: Energy released in conversion of 6  1023
atoms of Cl–
ions
= 6  1023
 electron affinity = 6  1023
 3.61 = 2.166  1024
eV
Let x Cl atoms are converted to Cl+
ion
Energy absorbed = x  ionization energy
x  17.422 = 2.166  1024
x = 1.243  1023
atoms
 (A)
Problem 19: The binding energy of an electron in the ground state of the He atom is equal to
24 eV. The energy required to remove both the electrons from the atom will be
(A) 59 eV (B) 81 eV
(C) 79 eV (D) None of these
Solution: Ionization energy of He
= 6
.
13
2
2

n
z
= 6
.
13
1
2
2
2

= 54.4 eV
Energy required to remove both the electrons
= binding energy + ionization energy
= 24. 6 + 54.4
= 79 eV
 (C)
Problem 20: The wave number of the shortest wave length transition in Balmer series of
atomic hydrogen will be :
(A) 4215Å (B) 1437 Å
(C) 3942 Å (D) 3647 Å
Solution:
shortest

1
= RZ2








 2
2
2
1
1
1
n
n
= 109678  12
 






 2
2
1
2
1
 = 3.647  10–5
cm
= 3647 Å
 (D)

43. 15. Assignments (Subjective problems)
LEVEL – I
1. a) What is wavelength of a particle of mass 1g moving with a velocity of 200 m/s?
b) A moving electron has 4.55 10-25
joules of kinetic energy. Calculate its
wavelength (mass = 9.1 10–31
kg and h = 6.610–34
kg m2
sec–1
)
c) Calculate the de-Broglie wavelength of electron accelerated through 100 volt.
2. The electron in a hydrogen atom revolves in the third orbit. Calculate (i) the energy of
the electron in this orbit (ii) the radius of the third orbit and (iii) the frequency and
wavelength of the spectral line emitted when the electron jumps from the third orbit to
the ground state.
(Given me = 9.1  10–28
g; e = 4.8  10–10
esu; h = 6.6  10–27
erg sec
3 Calculate the radii of the I, II and III permitted electron Bohr orbits in a hydrogen
atom. What are the corresponding values in the case of a singly ionised helium
atom?
4. Calculate energy in kcal/mole necessary to remove an electron in a hydrogen atom
in fourth principal quantum number of infinity.
5. An electron collides with a hydrogen atom in its ground state and excited it to a state
of n = 3. How much energy was given to the hydrogen atom in this inelastic
collision?
6. What is the wavelength of the first line is Paschen series of the
hydrogen spectrum?
7. Calculate the wavelength of first line of the (i) Lyman series and (ii) Balmer series,
assuming the Rydberg constant as 109, 678 cm–1
8. A monochromatic source of light operating at 600 watt emits 2 1022
photons per
second. Find the wavelength of the light.
9 a) The electron energy in hydrogen atom is given by E = – 21.7  10–12
/n2
erg.
Calculate the energy required to remove an electron completely from n = 2 orbit.
What is the longest wavelength (in cm) of light that can be used to cause this
transition?
b) Ionisation energy of hydrogen atom is 13.6 eV. Calculate the ionisation energy
for Li2+
and Be3+
in the first excited state.
10. A bulb emits light of wave length 4500Å. The bulb is rated as 150 watt and 8% of the
energy is emitted as light. How many photons are emitted by bulb per second.

44. LEVEL – II
1. a) What is accelerating potential needed to produce an electron beam with an
effective wavelength of 0.09Å.
b) Calculate de-Broglie wavelength of an electron moving with a speed of nearly
20
1
th that of light (3  108
ms–1
)
2 A doubly ionised lithium atom is hydrogen like with an atomic no. 3.
i) Find the wavelength of the radiation required to excite the electron in Li from the
first to the third Bohr orbit (ionisation energy of the hydrogen atom is equal to
13.6 eV)
ii) How many spectral lines are observed in the emission spectrum of the above
excited system?
3. a) Light from a discharge tube containing hydrogen atoms falls on the surface of a
piece of sodium. The kinetic energy of the fastest photo electrons emitted from
sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find
i) The energy of the photons causing the photo electric emission.
ii) The quantum numbers of the two levels involved in the emission of these
photons.
b) Find the threshold wavelengths for photoelectric effect from a copper surface, a
sodium surface and a cesium surface. The work function of these metals are 4.5
eV, 2.3 eV and 1.9 eV respectively.
c) Energy required to stop the ejection of electrons from Cu plate is 0.24 eV.
Calculate the work function when radiation of  = 253.7 nm strikes the plate?
4. An electron in order to have a wavelength of 500Å, through what potential difference
it must pass?
5. Calculate  of the radiations when the electron jumps from III to II orbit of hydrogen
atom. The electronic energy in II and III Bohr orbit of hydrogen atoms are
– 5.42 10–12
and –2.41 10–12
erg respectively.
6. What is uncertainty in velocity of an electron if uncertainty in its position is 1Å?
7. When a metal surface is irradiated by light of wave length 300 m; the stopping
potential is found to be 0.5V. Compute the work function and threshold wave length.
Also calculate the stopping potential required for light of wave length 200m.
8. The photoelectric effect consists of the emission of electron from the surface of the
metal when the metal is irradiated with light. A photon with a minimum energy of
3.97 10–19
J is necessary to eject an electron from barium.
a) What is frequency of the radiation corresponding to this value?
b) Will the blue light with wave length 450 nm be able to eject the electron?

45. 9. Find out the number of waves made by Bohr electron in one complete revolution in
its third orbit. Also calculate the number of revolutions per second that this electron
makes around the nucleus.
10. The dissociation of I2 
 
 
h 2I utilizes one photon per iodine molecule
dissociated. The maximum  for this is 4995 Å. Calculate number of moles of I2
dissociated per KJ of photon energy.

46. LEVEL – III
1. What is uncertainty in the location of a photon of wavelength 5000Å if wavelength is
known to an accuracy of 1 ppm.
2. A single electron beam, atom has nuclear charge +Ze where Z is atomic number and
e is electronic charge. It requires 16.52eV to excite the electron from the second
Bohr orbit to third Bohr orbit. Find
i) The atomic no. of element
ii) The energy required for transition of electron from first to third orbit.
iii) Wavelength required to remove electron from first Bohr orbit to infinity.
iv) The kinetic energy of electron is first Bohr orbit.
3. Calculate ionisational potential of positronium species consisting of an electron
bound to a positron.
4. A sample of hydrogen gas containing some atoms in one excited state emitted three
different types of photons. When the sample was exposed with radiation of energy
2.88 eV it emitted 10 different types of photons, all having energy equal or less than
13.05 eV. Find out
a) The principal quantum numbers of initially excited electrons.
b) The principal quantum numbers of electrons in final excited state.
c) The maximum and minimum energies of initially emitted photons.
5. A hydrogen like atom (atomic no. Z) is in a higher excited state of quantum number
‘n’. This excited atom can make a transition to the first excited state by successively
emitting two photons of energies 10.2 eV and 17.00 eV respectively. Alternatively,
the atom from the same excited state can make a transition to the second excited
state by successively emitting two photons of energy 4.25 eV and 5.95 eV
respectively. Determine the values of n and Z.
6. The photo electric emission requires a threshold frequency 0. For a certain metal 1
= 2200Å and 2 = 1900 Å, produce electrons with a maximum kinetic energy KE1 &
KE2. If E2 = 2KE1 calculate 0 and corresponding 0.
7. 1.8 g hydrogen atoms are excited to radiations. The study of spectra indicates that
27% of the atoms are in 3rd
energy level and 15% of atoms in 2nd
energy level and
the rest in ground state. IP of H is 13.6 eV. Calculate (i) No. of atoms present in III
and II energy level (ii) Total energy evolved when all the atoms return to ground
state.
8. The IP of H is 13.6 eV. It is exposed to electromagnetic waves of 1028 A and given
out induced radiations. Find the wavelength of these induced radiations.
9. The ionisation energy of a H like Bohr atoms is 4 Rydberg.
a) Calculate the wavelength radiated when electron jumps from the first excited
state to ground state.
b) What is radius of I orbit of this atom?

47. 10. Iodine molecular dissociates into atoms after absorbing light of 4500Å. If one
quantum of radiation is absorbed by each molecule, calculate the kinetic energy of
iodine atoms. Bond energy of I2 = 240 kJ mol–1
.
11. Two hydrogen atoms collide head on and end up with zero kinetic energy. Each then
emits a photon with a wavelength 121.6 nm. Which transition leads to this
wavelength. How fast were the hydrogen atoms travelling before the collision? Given
RH = 1.097  107
ms–1
& mH = 1.67  10–27
kg.
12. A mixture contains atoms of fluorine and chlorine. The removal of an electron from
each atoms of the sample absorbs 272.2 kJ while the addition of an electron to each
atom of the mixture releases 68.4 kJ. Determine the percentage composition of the
mixture, given that the ionisation energies of F & Cl are 27.91  10–22
& 20.77  10–22
kJ respectively and that the electron affinities are 5.53  10–22
and 5.78  10–22
respectively.
13. A particle of charge equal to that of an electron and mass 200 times the mass of an
electron moves in a circular orbital around a nucleus of charge +3e. Assuming that
the Bohr model of the atom is applicable to this system.
a) Derive the expression for radius of the nth Bohr orbit.
b) Find the value of ‘n’ for which the radius of the orbit is approximately the same as
that of the first Bohr orbit for the hydrogen atom and
c) Find the wavelength of the radiation emitted when the revolving particle jumps
from the third orbit to the first. Given RH = 1.097  10–7
m–1
14. The dye uriflarine, when dissolved in water has its maximum light absorption at
4530Å and its maximum fluorescence emission at 5080Å. The number of
fluorescence quanta is, on the average 53% of the number of quanta absorbed.
Using the wavelengths of maximum absorption and emission, what percentage of
absorbed energy is emitted as fluorescence?
15. The minimum energy necessary to overcome the attractive force between the
electron and the surface of silver metal is 7.52  10–19
J. What will be the maximum
kinetic energy of the electrons ejected from silver which is being irradiated with
ultraviolet light having a wavelength 360Å?

48. 16. Assignments (Objective problems)
LEVEL – I
1. The ratio of energy of the electron in ground state of the hydrogen to electron in first
excited state of He+
is
(A) 1:4 (B) 1:1
(C) 1:8 (D) 1:16
2. Bohr model can explain spectrum of
(A) the hydrogen atom only
(B) all elements
(C) any atomic or ionic species having one electron only
(D) the hydrogen molecule
3. Which is the correct order of probability of being found close to the nucleus is
(A) s  p  d  f (B) f  d  p  s
(C) p  d  f  s (D) d  f  p  s
4. Ratio between longest wavelength of H atom in Lyman series to the shortest
wavelength in Balmer series of He+
is
(A)
3
4
(B)
5
36
(C)
4
1
(D)
9
5
5. If the radius of first Bohr orbit is a, then de-Broglie wavelength of electron in 3rd
orbit
is nearly.
(A) 2a1 (B) 6a1
(C) 9a1 (D) 16a1
6. If uncertainty in position and momentum are equal, the uncertainty in velocity would
be
(A)

h
m
2
1
(B)

2
h
(C)

h
(D)
m
h

7. Which of the following is a coloured ion
(A) Cu+
(aq) (B) Na+
(aq)
(C) Cu2+
(aq) (D) K+
(aq)
8. The number of orbitals in a sub-shell are given by
(A) 2l (B) n2
(C) 2l + 1 (D) 2n2
9. For a ‘d’ electron, the orbital angular momentum is
(A) 
6 (B) 
2
(C)  (D) 
2

49. 10. Hydrogen atom consists of a single electron but so many lines appear in the
spectrum of atomic hydrogen because
(A) Sample contains some impurity
(B) Experiment is done on collection of atoms.
(C) Hydrogen atom splits to form more than one different species.
(D) Some different isotope of hydrogen atom may be present.
11. If the radius of first Bohr orbit is x, then de Broglie wavelength of electron in 3rd
orbit
is nearly.
(A) 2x (B) 6x
(C) 9x (D)
3
x
12. One molecule of a substance absorbs one quantum of energy. The energy involved
when 1.5 mole of the substance absorbs red light of frequency
7.5 1014
sec–1
will be
(A) 2.99 105
J (B) 3.23 105
J
(C) 4.48 105
J (D) 2.99 106
J
13. If a shell is having g sub-shell, which is correct statement about principal quantum
number n of this shell.
(A) n  5 (B) n  5
(C) n = 5 (D) Cannot be determined
14. The wave number of a spectral line is 5 105
m–1
. The energy corresponding to this
line will be
(A) 3.3910–23
kJ (B) 9.93 10–23
kJ
(C) 3.45 10–24
J (D) none of these
15. Wave length of the radiation when electron jumps from second shell to 1st
shell of H
atom (RH = 109679 cm–1
)
(A) 1215.6Å (B) 1397.5Å
(C) 2395.87 Å (D) none of these

50. LEVEL – II
1. Which statement is wrong about Bohr’s theory
(A) Orbit is a three dimensional area where probability of finding electron is
maximum.
(B) Orbit is a two dimensional track on which electron moves
(C) Atom has definite boundary
(D) Energies and angular momentum of orbits are quantized.
2. The first five ionization energies of an element are 801, 2428, 3660, 25030, 32835 in
kJ/mol. Then the element could be
(A) a halogen (B) a noble gas
(C) a third group element (D) a second group element
3. Which statement is true
(A) Spacing between energy levels n = 1 and n = 2 in hydrogen atom is greater than
that of n = 2 and n = 3
(B) Spacing between energy levels n = 1 and n = 2 in hydrogen atom is equal to that
n = 2 and n = 3
(C) Spacing between energy levels n = 1 and n = 3 in hydrogen atom is less than
that of n = 2 and n = 3
(D) None
4. If the wave length of first line of the Balmer series of hydrogen atom is 656.1 nm, the
wave length of second line of this series would be
(A) 218.7 nm (B) 328.0 nm
(C) 486.0 nm (D) 640.0 nm
5. In presence of external magnetic field f sub-shell is
(A) 5 fold degenerate (B) 3 fold degenerate
(C) 7 fold degenerate (D) non-degenerate
6. Which set of quantum numbers is not possible for electron in 3rd
shell?
(A) n = 3 l = 2 m = – 1 s = +1/2
(B) n = 3 l = 2 m = – 1 s = –1/2
(C) n = 3 l = 2 m = 0 s = +1/2
(D) n = 3 l = 3 m = 0 s = –1/2
7. The first four ionization energies of an element are 191, 578,872 and 5962 kcal. The
number of valence electrons in the element is
(A) 1 (B) 2
(C) 3 (D) 4
8. The radiation is emitted when a hydrogen atom goes from a high energy state to a
lower energy state. The wavelength of one line in visible region of atomic spectrum
of hydrogen is 6.5  10–9
m. Energy difference between the two states is
(A) 3.0 10–17
J (B) 1.0 10–18
J
(C) 5.0  10–10
J (D) 6.5 10–7
J

51. 9. The ratio of the energy of the electron in ground state of hydrogen to the electron in
first excited state of Be3+
is
(A) 1 : 4 (B) 1 : 8
(C) 1 : 16 (D) 16 : 1
10. Which one of the following species is isoelectronic with P3–
?
(A) Kr (B) Ca2+
(C) Na+
(D) F–
11. Which of the following sets of quantum numbers is/are not allowable.
(A) n = 3 l = 2 m = 0
(B) n = 2 l = 0 m = – 1
(C) n = 4 l = 3 m = + 1
(D) n = 1 l = 0 m = 0
12. Among V (Z = 23), Cr (Z = 24), Mn (Z = 25) which will have highest magnetic
moment.
(A) V (B) Cr
(C) Mn (D) all of them will have equal magnetic
moment
13. If velocity of an electron in 1st
Bohr orbit of hydrogen atom is x, its velocity in 3rd
orbit
will be
(A)
3
x
(B) 3x
(C) 9x (D)
9
x
14. Which element has a hydrogen like spectrum whose lines have wavelengths one
fourth of atomic hydrogen?
(A) He+
(B) Li 2+
(C) Be3+
(D) B4+
15. Calculate the wavelength of a track star running 150 metre dash in 12.1 sec if its
weight is 50 kg.
(A) 9.11  10–34
m (B) 8.92  10–37
m
(C) 1.12  10–45
metre (D) none of these

52. 17. Answers to Subjective Assignments
LEVEL – I
1. a) 3.313  10–33m 2. i) – 2.43  10–12 erg
b) 7.25  10–7m ii) 4.75Å
c) 1.227Å iii) 19.44  10–12erg, 2.95  1015Hz. 1017Å
3. r1= 0.53Å r1 = 0.265Å
r2 = 2.12Å For hydrogen r2 = 1.06Å For He+
r3 = 4.77Å r3 = 2.385Å
4. 19.5 kcal/mole
5. 12.08 eV 6. 1.875  10–4 cm
7. 1215Å and 6565Å 8. 6630 nm 9.a) 3.67  10–5 cm
10. 27.2  1018 b) 30.6 eV for Li2+, 54.4 eV for Be3+
LEVEL – II
1. a) 1.86  104 eV 2. i) 113.7Å
b) 4.85  10–11m ii) 3
3. a) i) 2.55eV 4. 6.03  10–4 eV
ii) 2  4 5. 6603Å
b) 276 mm, 540 nm, 654 nm 6. 5.79  105
c) 4.65 eV
7. 2.57 volt 8. 5.99  1014 sec–1 yrs
9. 3 and 2.43  1014 10. 4.17  10–3
LEVEL – III
1. 4  10–2m 2. i) 3, (ii) 108.87 eV, (iii) 1.01  10–6cm
3. 6.8 eV iv) 122.4 eV
4. a) 2 and 3 5. n = 6 and z = 3
b) 5 6. 2612.6Å
c) 12.09 eV and 1.9 eV 9. a) 3.03.89Å
7. i) 628.72  1021 atoms b) 2.645  10–9 cm
ii) 832.50 kJ. 10. 0.216  10–19J
8. 1 = 1028Å, 2= 1216Å, 3 = 6558Å
11. n = 1 to n = 2, u = 4.42  104 msec–1 12. % of fluorine atoms 27.27
13. a) r = n2 






 )
4
/
Ze
(
m
4
h
0
2
2
2
14.
1
2
n
n
= 0.59
b) n = 24 15. 47.68  10–19 Joule
c) 57.76  10–12 m

53. 18. Answers to Objective Assignments
LEVEL – I
1. B 2. B
3. A 4. A
5. B 6. A
7. C 8. C
9. A 10. B
11. B 12. C
13. B 14. B
15. A
LEVEL – II
1. A 2. C
3. A 4. C
5. C 6. D
7. C 8. A
9. A 10. B
11. B 12. B
13. A 14. A
15. B
6